characterization of field
Proposition 1.
Let be a commutative ring with identity. The ring (as above) is a field if and only if
has exactly two ideals: .
Proof.
() Suppose is a field and let be a non-zero ideal of . Then there exists with . Since is a field and is a non-zero element, there exists such that
Moreover, is an ideal, , so . Hence . We have proved that the only ideals of are and as desired.
() Suppose the ring has only two
ideals, namely . Let be a
non-zero element; we would like to prove the existence of a
multiplicative inverse for in . Define the
following set:
This is clearly an ideal, the ideal
generated by the element . Moreover, this ideal is not the zero
ideal because and was assumed to be
non-zero. Thus, since there are only two ideals, we conclude
. Therefore so there exists an element such that
Hence for all non-zero , has a multiplicative inverse in , so is, in fact, a field. โ
Title | characterization of field |
---|---|
Canonical name | CharacterizationOfField |
Date of creation | 2013-03-22 13:57:03 |
Last modified on | 2013-03-22 13:57:03 |
Owner | alozano (2414) |
Last modified by | alozano (2414) |
Numerical id | 7 |
Author | alozano (2414) |
Entry type | Theorem |
Classification | msc 12E99 |
Synonym | a field only has two ideals |
Related topic | Field |
Related topic | Ring |
Related topic | Ideal |